Problem 8 - Thomas Calculus in SI Units

Question

Which of the functions satisfy the hypotheses of the Mean Value Theorem on the given interval, and which do not? Give reasons for your answers. $$f(x)=x^{4 / 5}, \quad[0,1]$$

Step-by-Step Solution

Answer

The function does not satisfy the Mean Value Theorem on \([0, 1]\) because it is not differentiable at \( x = 0 \).

Step 1: Understanding the Mean Value Theorem

The Mean Value Theorem states that if a function \( f \) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one number \( c \) in \((a, b)\) such that \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]

Step 2: Checking Continuity on [0, 1]

The function \( f(x) = x^{4/5} \) is a power function where the exponent \( \frac{4}{5} \) is a positive rational number. Power functions are continuous everywhere where their base is positive. Therefore, \( f(x) = x^{4/5} \) is continuous on the interval \([0, 1]\) since there are no points where the function is undefined or discontinuous.

Step 3: Checking Differentiability on (0, 1)

To check differentiability, we find the derivative of \( f(x) \). The derivative is \[ f'(x) = \frac{4}{5}x^{-1/5}. \]The function \( f'(x) \) is defined and continuous for all \( x > 0 \). Thus, \( f(x) \) is differentiable on \((0, 1)\) because there are no discontinuities or undefined points on this interval.

Step 4: Checking Differentiability at x = 0

At \( x = 0 \), the derivative \( f'(x) = \frac{4}{5}x^{-1/5} \) becomes undefined since division by zero is not possible. Therefore, \( f(x) \) is not differentiable at \( x = 0 \).

Step 5: Conclusion

Since \( f(x) = x^{4/5} \) is continuous on \([0, 1]\) but not differentiable at \( x = 0 \), the hypotheses of the Mean Value Theorem are not fully satisfied because the function fails to be differentiable over the entire open interval \((0, 1)\).

Key Concepts

Continuity

Continuity is a core concept in calculus that forms the foundation for many theorems and principles, including the Mean Value Theorem (MVT). A function is said to be continuous over a certain interval if there are no breaks, jumps, or holes in its graph within that interval.

For the function \( f(x) = x^{4/5} \), which is a power function, continuity must be assessed on the given interval \([0, 1]\). Fortunately, power functions with positive rational exponents like \( \frac{4}{5} \) are continuous over intervals where their domain is defined without restrictions, specifically where the base \( x \geq 0 \).

• The function \( f(x) = x^{4/5} \) remains connected throughout the interval \([0, 1]\).
• There are no interruptions in its graph, indicating a smooth path between any two points in this range.

Therefore, we can affirm that \( f(x) \) is continuous over \([0, 1]\). However, ensuring continuity alone isn’t sufficient for MVT; differentiability must also be verified.

Differentiability

Differentiability refers to the existence of a derivative at each point in a function's domain. It's a step above continuity, suggesting not only that a function is unbroken but also that it has a well-defined tangent line at each point in the specified interval.

For the function \( f(x) = x^{4/5} \), we find that the derivative is \( f'(x) = \frac{4}{5}x^{-1/5} \). This expression is continuous and defined for \( x > 0 \). As such, \( f(x) \) is differentiable everywhere on the interval \((0, 1)\), meaning the function behaves predictably with smoothly varying slopes.

• Inside \((0, 1)\), \( f'(x) \) does not encounter any undefined values or abrupt changes.
• The derivative only presents issues at the boundary point \( x = 0 \).

At \( x = 0 \), \( f'(x) \) becomes undefined due to division by zero (\( x^{-1/5} \) involves a negative exponent). Thus, "differentiability at endpoints" is a crucial check, and the lack of a defined derivative at \( x = 0 \) implies \( f(x) \) is not differentiable at this point. This fact impacts satisfying the hypotheses of the MVT.

Power Function

A power function is a type of polynomial function characterized by a base and an exponent, usually in the form \( f(x) = x^n \), where \( n \) is a constant. Power functions can exhibit a broad range of behaviors, influenced primarily by the value of \( n \).

• For integers and positive rational numbers as exponents, they tend to be continuous and often differentiable, except possibly at points concerning boundary conditions.
• The function \( f(x) = x^{4/5} \) particularly shows these characteristics on fairly simple intervals like \([0, 1]\).

What makes power functions like \( x^{4/5} \) special in calculations involving the Mean Value Theorem is their predictability and ease of differentiation, albeit with the nontrivial necessity to confirm endpoints. In this case, while the function retains continuity on \([0, 1]\), issues arise at \( x = 0 \) concerning differentiability. Understanding these nuances is key to applying MVT in calculus comprehensively.